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erik [133]
3 years ago
10

Your brother, who is prone to bearing substantial risk, suggests that you buy a security for $10,000 that promises to pay you $1

00,000 at the end of 15 years. What is the implied annual return or yield on this investment
Business
1 answer:
astraxan [27]3 years ago
5 0

Answer:

16.59%

Explanation:

First we look at the formula which to determine the future value of the security and then work back to determine the annual return in terms of percentage

Future Value = Present Value x (1 +i)∧n

where i = the annual rate of return

n= number of years or period

We then plug the given figures into the equation as follows

we already know Present value to be $10,000 and the future value to be $100,000 and the number of years to be 15

Therefore, the implied annual return or yield on the investment is

100,000 = 10,000 x (1+i)∧15

(1+i)∧15 = 100,000/10,000 = 10

1 + i = (10∧(1/15))=1.165914

i= 1.165914-1

= 0.1659

= 16.59%

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You want to take a dream vacation in 3.5 years. You plan to save up $5,000 in your vacation sinking fund. Assume an interest rat
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<h3>Answer:</h3>

<h3>Explanation:</h3>

The formula for calculating the Monthly payments P for the sinking fund is as follows:

P\;=\;\frac{A*i}{(1+i)^n-1}

where,

P = Monthly payments to be made

A = Total amount to be accumulated

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Assuming interest is applied at the beginning of each period.

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In this scenario, as some of the year is already passed (assume 6 months), to complete the time period of 3.5 years the interest will compound 3 times (as the 0.5 year payments can be adjusted in the remaining part of the first year and no interest is applied on it). Hence, the interest will be applied 3 times.

\therefore P_{(i)}\;=\;\frac{5000*0.08}{(1+0.08)^3-1}\\\\P_{(i)}\;=\;\frac{400}{0.2597}\\\\P_{(i)}\;=\;1540.1676

<h3><u>Scenario (ii) - Deposit is made at the beginning of the year:</u></h3>

For this case, the interest will be applied 4 times to complete the time period of 3.5 years for payment.

\therefore P_{(ii)}\;=\;\frac{5000*0.08}{(1+0.08)^4-1}\\\\P_{(ii)}\;=\;\frac{400}{0.3605}\\\\P_{(ii)}\;=\;1109.6040

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